p. 1 Math 490 Notes 20 More About Compactness Recall that in Munkres it is proved that a simply (totally) ordered set X with the order topology is connected iff it satisfies: (1) Every subset bounded above has a least upper bound; (2) If x < y, there is a z such that x < z < y. Furthermore, if X is connected, then any interval in X is also connected. In particular, this theorem implies that every interval in R is connected, and we ve showed that any non-interval in R is disconnected. Thm 27.1 in Munkres is a related result pertaining to compactness in simply ordered sets. We ll state only the result here, since the proof is given in Munkres. Prop N20.1 If X is a simply ordered set with the order topology which has the l.u.b property (see (1) above), then every closed interval [a,b] in X is compact. Note: For a closed interval [a,b] in a simply ordered set, conditions (1) and (2) above are necessary and sufficient for connectedness, whereas only (1) is necessary and sufficient for compactness. Heine-Borel Theorem A subset of (R,τ u ) is compact iff it is closed and bounded. Proof : By Prop N20.1, any closed, bounded interval [a, b] in R is compact. Any closed, bounded set in R is a closed subset of some interval [a,b], and hence compact by Prop N18.2(a). Conversely, any compact set A in R is closed by Prop N18.2(b). Also, A must be bounded since an unbounded set in any metric space can be covered by a collection of bounded open balls, and such a cover has no finite subcover.
p. 2 Generalized Heine-Borel Theorem A subset of (R n,τ u ) is compact iff it is closed and bounded. Proof : A compact set in any metric space must be closed and bounded, as shown in the preceding proof. Conversely, any closed, bounded set in R n is a closed subset of a product of closed, bounded intervals in R, and the latter is compact by the Tychanov Theorem. A standard example: the Long Line Let S Ω [0, 1) have the dictionary order topology, and let L be the subspace obtained by deleting the least element (0, 0). We may visualize L as an uncountable number of vertical columns, each (except for the first) a copy of [0, 1), indexed by the countable ordinals. L is a linear continuum, and hence connected. How is L related to (R,τ u )? One may show that L is locally like R in the sense that each point in L has a nbhd basis of sets homeomorphic to open intervals in R (this is not so obvious for points of the form (0,λ) L, where λ is a limit ordinal). Thus L, like (R,τ u ), is first countable. However L is not second countable, since L contains uncountably many disjoint open sets (choose an open interval in each column). Thus (R,τ u ) is homeomorphic to a subspace of L (indeed, to the first column), but L is not homeomorphic to a subspace of R, since second countability is hereditary. Also, L is not metrizable, since it contains a subspace homeomorphic to S Ω, which is not metrizable (we re not ready to prove this yet).
p. 3 Compactness in Metric Spaces For metric spaces, there are several ways to characterize compactness which are non-equivalent when generalized to topological spaces. Three such alternative versions of compactness (which are equivalent in metric spaces, but not in topological spaces) are defined as follows: Def N20.1 A topological space (X,τ) is: (1) limit-point compact iff each infinite subset of X has a limit point; (2) countably compact iff each countable open cover of X has a finite subcover; (3) sequentially compact iff every sequence in X has a convergent subsequence. Prop N20.2 In any toplogical space, (compact) (countably compact) (limit-point compact). For a T 1 space, (limit-point compact) (countably compact). Proof : Exercise. Examples Example 2 in 28 of Munkres shows that S Ω (in the order topology) is limit-point compact, and hence countably compact, since it is T 2. One may also show that S Ω is sequentially compact, since every sequence in S Ω has a subsequence which is either constant or increasing, and in either case convergent. Since S Ω is not compact, none of these properties implies compactness. Furthermore, compactness does not imply sequential compactness, since [0, 1] [0,1] with the product topology is compact by the Tychanov theorem, but it can be shown not to be sequentially compact. More results about compactness: Prop N20.3 ((X,τ) sequentially compact) ((X,τ) countably compact). Prop N20.4 If (X,τ) is T 1 and first-countable, then ((X,τ) limit-point compact) ((X,τ) sequentially compact).
p. 4 Theorem N20.1 In a metric space (X, τ), (compact) (sequentially compact) (limitpoint compact) (countably compact). Note: The proof of Theorem N20.1 requires the use of the following well-known lemma, which is proved on p.175 of Munkres: Lebesque Covering Lemma If (X, d) is a metric space which is sequentially compact and U is an open cover of X, then there exists a δ > 0 (called the Lebesque number for U) such that if A X and and diama def = sup{d(x,y) x,y A} < δ, then there is a U U such that A U. Recall that a function f : (X,d) (Y,ρ) between two metric spaces is said to be uniformly continuous iff for every ǫ > 0, there exists a δ > 0 such that d(x,y) < δ ρ(f(x),f(y)) < ǫ. Another possibly familiar result whose proof (see Munkres p. 176) uses the Lebesque Covering Lemma is: Uniform Continuity Theorem: If f : (X,τ) (Y,µ) is continuous and (X,τ) is compact, then f is uniformly continuous. Example An interesting property of the ordinal line S Ω is that any continuous map f :S Ω (R,τ u ) is eventually constant, meaning that there exists λ S Ω such that f(µ) = f(λ) for all µ > λ. The proof of this result is based on the following facts: (1) S Ω is countably compact; (2) Countable compactness is preserved under continuous maps; (3) f(s Ω ) is compact in (R,τ u ) by Thm 1. We ll omit the remaining details.
p. 5 Prop N20.5 A compact, metrizable space is separable. Proof : Let (X,τ) be compact and metrizable, with d being a metric which induces τ. By compactness, for any n N, the set of all open balls of radius 1 n has a finite subcover B n = {B(x 1n, 1),B(x n 2 n, 1,...,B(x n k n, 1 )}. Note that for each n N, there is a different, n finite set of center points {x 1n,x 2n,...,x kn }. Let D be the countable collection of all these sets of center points: D = n N{x 1n,x 2n,...,x kn }. The claim now is that D is a dense subset of X. To see this, let x X and U a nbhd of x. Then there exists an ǫ > 0 such that B(x,ǫ) U. Choose m N such that 1 m < ǫ. Since {B(x 1m, 1 ),B(x m 2 m, 1,...,B(x m k m, 1 )} is a cover of X, we know that x B(x m j m, 1 ) for some m x jm. But this implies that d(x,x jm ) < ǫ, so B(x,ǫ) D φ, so U D φ. Corollary For metrizable spaces: (compact) (sequentiall compact) (limit point compact) countably compact) (separable) (2nd countable).